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| 1. |
Find the value of cos60 geometrically. |
| Answer» Let a triangle ABC with each side equal to 2a{tex}\\angle{/tex}A = {tex}\\angle{/tex}B = {tex}\\angle{/tex}C = 60°Draw AD perpendicular to BCIn\xa0{tex}\\triangle{/tex}BDA and\xa0{tex}\\triangle{/tex}CDAAB = AC, AD = AD and\xa0{tex}\\angle BDA=\\angle CDA{/tex}{tex}\\triangle B D A \\cong \\triangle C D A{/tex}\xa0by RHSBD = CD{tex}\\angle{/tex}BAD = {tex}\\angle{/tex}CAD = 30° by CPCT{tex}\\cos 60 ^ { \\circ } =\\frac BH= \\frac { B D } { A B } = \\frac { a } { 2 a } = \\frac { 1 } { 2 }{/tex} | |